What is the vertex of #y=2x^2+6x+4#?

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Answer 1 · 2018-06-13T20:43:58

#V = (-3/2, - 1/2)#

#V = (-b/(2a), - Delta/(4a))#

#Delta = 36 - 4 * 2 * 4 = 4#

#V = (-6/4, - 4/8)#

Answer 2 · 2018-06-13T20:45:39

#(-\frac{3}{2},-\frac{1}{2})#

Method 1: Calculus approach

Vertex is where the gradient of the curve is 0.

Therefore find #\frac{dy}{dx}#

#\frac{dy}{dx}=4x+6#

Equate that to 0 such that:

#4x+6=0#

Solve for #x#, #x=-\frac{3}{2}#

Let #x=-\frac{3}{2}# into the original function, therefore

#y=2*(-\frac{3}{2})^{2}+6*(-\frac{3}{2})+4#

#y=-\frac{1}{2}#

Method 2: Algebraic approach.

Complete the square to find the turning points, also known as the vertex.

#y=2x^{2}+6x+4#
#y=2(x^{2}+3x+2)#
#y=2[(x+\frac{3}{2})^{2}-\frac{9}{3}+2]#
#y=2(x+\frac{3}{2})^{2}-\frac{1}{2}#

Notice here that you have to multiply BOTH terms by 2, as 2 was the common factor which you took out of the entire expression!

Therefore, the turning points can be picked up such that
#x=-\frac{3}{2},y=-\frac{1}{2}#

Therefore coordinates:

#(-\frac{3}{2},-\frac{1}{2})#